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Oh, and I have to write the first four terms...
So to sum up, the sequence is neither arithmetic nor geometric
Wouldn't term 2 be 5/2?
oh right, made a typo and mixed up d and the second term you are correct, T2 = 2+1/2 = 5/2
Ok, could you check term one as well? I'm getting 2.
yeah another very silly typo, 1+1/1 = 1+1 = 2...lol not sure how i missed that...
The first four terms are T1 = 1+1/1 = 2 T2 = 2+1/2 = 5/2 T3 = 3+1/3 = 10/3 T4 = 4+1/4 = 17/4 So the first four terms are T1 = 2 T2 = 5/2 T3 = 10/3 T4 = 17/4
I have another problem like this one... t sub n = 16 times 2^2n Same thing, figure out what kind it is, if any, and give the first four terms.
T1 = 16*2^(2*1) = 16*2^2 = 16*4 = 64 T2 = 16*2^(2*2) = 16*2^4 = 16*16 = 256 T3 = 16*2^(2*3) = 16*2^6 = 16*64 = 1024 T4 = 16*2^(2*4) = 16*2^8 = 16*256 = 4096 So the first four terms are T1 = 64 T2 = 256 T3 = 1024 T4 = 4096 My initial guess is that this is a geometric sequence (since we have an exponential expression) If this is case, then the following must be true r = T2/T1 r = T3/T2 r = T4/T3 where r is some number that is the same for all 3 equations So r = T2/T1 = 256/64 = 4, which makes r = 4 r = T3/T2 = 1024/256 = 4, which makes r = 4, so far so good, last one r = T4/T3 = 4096/1024 = 4, which makes r = 4 The common ratio is r = 4 every time. So this is indeed a geometric sequence. Alternatively, we can write T[n] = 16*2^(2n) into the form T[n] = 16 * 2^n * 2^n T[n] = 16*(2*2)^n T[n] = 16*(4)^n and we can say that the last equation shown above is in the form T[n] = a*r^n where r = 4. This further confirms that this is a geometric sequence.